More Algebra questions

[Zap Brannigan]

Ok now I need to evaluate the square root of 144 +(-27)^1/3

 

[Zap Brannigan]

I got a solution of 9

 

[Zap Brannigan]

Now I need to evaluate the square root of 144a^6 b^24

 

[hypn0tik]

If cube root is exponentiating to the power of 1/3, square is exponentiating to the power of 1/2. The nth root is taking the exponent of 1/n.

 

[Zap Brannigan]

I see the 12 but do I divide the exponents by 2?

 

[Zap Brannigan]

Originally posted by: hypn0tik
If cube root is exponentiating to the power of 1/3, square is exponentiating to the power of 1/2. The nth root is taking the exponent of 1/n.

Just read this time to rework the problem...

 

[Zap Brannigan]

Is it 12a^3 b^12?

 

[hypn0tik]

Originally posted by: Zap Brannigan
Is it 12a^3 b^12?

Yep.

 

[Zap Brannigan]

Yes it must be!

 

[Zap Brannigan]

This is fun! I know a ^3 is cubed root but what is a ^4 called?

Going to rewrite this next problem right off as ( a^12 b^4)^1/4

 

[Zap Brannigan]

Which brings the solution of a^3 b

 

[hypn0tik]

Originally posted by: Zap Brannigan
This is fun! I know a ^3 is cubed root but what is a ^4 called?

Going to rewrite this next problem right off as ( a^12 b^4)^1/4

There's no special term. Simply the 'fourth root'.

 

[Zap Brannigan]

Ok now I have to replace the radical with a rational Exponent.

(5th root 2x^3)^3

Did I write that one correctly?

 

[hypn0tik]

Originally posted by: Zap Brannigan
Ok now I have to replace the radical with a rational Exponent.

(^5 root 2x^3)^3

Did I write that one correctly?

Nope. That doesn't make any sense.

 

[The Godfather]

messenger.msn.com

 

[Zap Brannigan]

Ok well it looks like this,

( a small 5 sitting on top of the v part of the square root sign, then 2x^3 closed parethesis) ^3

 

[Zap Brannigan]

I'm thinking it wants 1/5 of 2x^3 first?

 

[hypn0tik]

Originally posted by: Zap Brannigan
I'm thinking it wants 1/5 of 2x^3 first?

When you're exponentiating, it doesn't matter if you decide to cube first and then find the fifth root or find the fifth root first and then cube.

So your problem can be written as:

(2x^3)^3/5

 

[Zap Brannigan]

Still does'nt look easily solvable....

 

[hypn0tik]

Originally posted by: Zap Brannigan
Still does'nt look easily solvable....

Only difference here is you don't get pretty numbers.

 

[Zap Brannigan]

What happens to the 3 after dividing by 5?

 

[Howard]

(2x^3)^3/5 = 2^(3/5) * x^[3*(3/5)] (I think)

3*(3/5) is 9/5

 

[hypn0tik]

Originally posted by: Howard
(2x^3)^3/5 = 2^(3/5) * x^[3*(3/5)] (I think)

3*(3/5) is 9/5

Yes, your result is:

2^(3/5) x^(9/5)

2^(3/5) is something you would punch into your calculator.

 

[Zap Brannigan]

How would I punch that into my calculator? This is all new to me.

 

[hypn0tik]

Originally posted by: Zap Brannigan
How would I punch that into my calculator? This is all new to me.

Depends on your calculator. Look for it's instruction manual.